Complex roots of irreducible cubic in $\mathbb{Q}[x]$ Let $$f(x) = x^3 +ax^2 + cx + d \in \mathbb{Q}[x] $$ with one real root, and two complex roots: α and β
α and β are conjugates.
My task is to show that:
$$β \notin \mathbb{Q}(α)$$
I'm confused as I believed that the rational numbers extended by this complex number would contain its conjugate.
 A: If the conjugate is in there, then the polynomial has two roots, but then it must have all three, since $f(x) = (x-\alpha)(x-\beta)(x-\gamma)$ for some real number $\gamma$, and $\Bbb Q(\alpha)\cong K=\Bbb Q[x]/(f(x))$, which has degree $3$ over $\Bbb Q$ would contain all three roots. However, this is impossible as

$$K\cong \Bbb Q(\gamma)=\{a+b\gamma+c\gamma^2: a,b,c\in\Bbb Q\}$$

This field is completely contained within $\Bbb R$, so it contains no non-real numbers, i.e. it cannot have all three roots to the polynomial within itself. So it must be that $\beta\not\in\Bbb Q(\alpha)$.
A: Suppose $\beta \in \mathbb{Q}(\alpha) =: k$, then
$$
(x-\alpha)(x-\beta) \mid f(x)
$$
in $k[x]$. Thus, the real root $\gamma$ of $f$ is also in $k$. However, $[\mathbb{Q}(\gamma):\mathbb{Q}] = 3$ and $[k:\mathbb{Q}] = 3$, so $\mathbb{Q}(\gamma) = k$, and so all the root of $f$ are real, which is a contradiction.
A: Here is my attempt. Let the real root be $\gamma$. Since $f(x)$ is irreducible in $\mathbb{Q}[x]$, this $\gamma$ must be irrational. And Also $[\mathbb Q (\gamma):\mathbb Q] = 3$ and $[\mathbb Q(\alpha) : \mathbb Q] = 3$. Since $\alpha , \beta $ are complex, they are not in $\mathbb Q (\gamma) $ . So by thinking in $\mathbb{Q}(\gamma)[x]$, we get that $x^2 - (\alpha + \beta)x + \alpha \beta$ is irreducible in $\mathbb{Q}(\gamma)[x]$ . This implies $[\mathbb Q (\gamma, \alpha) : \mathbb Q(\gamma)] = 2$. So We have  $[\mathbb Q(\gamma, \alpha):\mathbb Q] =[\mathbb Q (\gamma, \alpha) : \mathbb Q(\gamma)][\mathbb Q(\gamma) : \mathbb Q] = 6 $. But note that, $\mathbb Q (\gamma, \alpha) = \mathbb Q(\alpha, \beta) $, which implies $[\mathbb Q(\alpha, \beta):\mathbb Q] = 6$. 
Now, if $\beta \in \mathbb Q (\alpha)$, we get $\mathbb Q(\alpha, \beta ) = \mathbb Q (\alpha) $. This implies the degree of $\mathbb Q (\alpha )$ over $\mathbb Q $ is 6 , which is a contradiction. 
A: As the sum of the roots is rational, having two roots (i.e., $\alpha, \beta\in \mathbf{Q}[\alpha]$)in a field would imply that it has the third root too. So we get a contradiction: that it is a field contained  reals, and that it contains a complex root.
A: As it stands, I don't think the question is quite correct, we need to add that $f(x)$ is irreducible or something?  If not then $x^3-1=0$ has one real root, $1$, and two complex roots, $\alpha = \frac{-1+i\sqrt3}{2}$ and $\beta=\frac{-1-i\sqrt3}{2}$ with $\beta\in \Bbb Q (\alpha)$ as $\beta = \alpha ^2$. Is that right?
I've just noticed that the title does say 'irreducible'.  Maybe worth editing the question?
